Diagram of a Round Town

The monography begins with a master diagram called the Diagram of Round Town(圆城图式). It shows a circle inscribed in a right angle triangle and four horizontal lines, four vertical lines.

• TLQ, the large right angle triangle, with horizontal line LQ, vertical line TQ and hypotenuse TL

C: Center of circle:

• NCS: A vertical line through C, intersect the circle and line LQ at N(南north side of city wall), intersects south side of circle at S(南).
• NCSR, Extension of line NCS to intersect hypotenuse TL at R(日)
• WCE: a horizontal line passing center C, intersects circle and line TQ at W(西, west side of city wall) and circle at E (东, east side of city wall).
• WCEB:extension of line WCE to intersect hypotenuse at B(川)
• KSYV: a horizontal tangent at S, intersects line TQ at K(坤), hypotenuse TL at Y(月).
• HEMV: vertical tangent of circle at point E, intersects line LQ at H, hypotenuse at M(山, mountain)
• HSYY, KSYV, HNQ, QSK form a square, with inscribed circle C.
• Line YS, vertical line from Y intersects line LQ at S(泉, spring)
• Line BJ, vertical line from point B, intersects line LQ at J(夕, night)
• RD, a horizontal line from R, intersects line TQ at D(旦, day)

The North, South, East and West direction in Li Zhi's diagram are opposite to our present convention.

Triangles and their sides

There are a total of fifteen right angle triangles formed by the intersection between triangle TLQ, the four horizontal lines, and four vertical lines.

The names of these right angle triangles and their sides are summarized in the following table

More information 天地乾 ...

Number	Name	Vertices	Hypotenuse0c	Vertical0b	Horizontal0a
1	通 TONG	天地乾 ${\displaystyle \triangle TLQ}$	通弦（TL天地）	通股（TQ天乾）	通勾（LQ地乾）
2	边 BIAN	天西川 ${\displaystyle \triangle TWB}$	边弦（TB天川）	边股（TW天西）	边勾（WB西川）
3	底 DI	日地北 ${\displaystyle \triangle RDN}$	底弦（RL日地）	底股（RN日北）	底勾（LB地北）
4	黄广 HUANGGUANG	天山金 ${\displaystyle \triangle TMJ}$	黄广弦（TM天山）	黄广股（TJ天金）	黄广勾（MJ山金）
5	黄长 HUANGCHANG	月地泉 ${\displaystyle \triangle YLS}$	黄长弦（YL月地）	黄长股（YS月泉）	黄长勾（LS地泉）
6	上高 SHANGGAO	天日旦 ${\displaystyle \triangle TRD}$	上高弦（TR天日）	上高股（TD天旦）	上高勾（RD日旦）
7	下高 XIAGAO	日山朱 ${\displaystyle \triangle RMZ}$	下高弦（RM日山）	下高股（RZ日朱）	下高勾（MZ山朱）
8	上平 SHANGPING	月川青 ${\displaystyle \triangle YSG}$	上平弦（YS月川）	上平股（YG月青）	上平勾（SG川青）
9	下平 XIAPING	川地夕 ${\displaystyle \triangle BLJ}$	下平弦（BL川地）	下平股（BJ川夕）	下平勾（LJ地夕）
10	大差 DACHA	天月坤 ${\displaystyle \triangle TYK}$	大差弦（TY天月）	大差股（TK天坤）	大差勾（YK月坤）
11	小差 XIAOCHA	山地艮 ${\displaystyle \triangle MLH}$	小差弦（ML山地）	小差股（MH山艮）	小差勾（LH地艮）
12	皇极 HUANGJI	日川心 ${\displaystyle \triangle RSC}$	皇极弦（RS日川）	皇极股（RC日心）	皇极勾（SC川心）
13	太虚 TAIXU	月山泛 ${\displaystyle \triangle YMF}$	太虚弦（YM月山）	太虚股（YF月泛）	太虚勾（MF山泛）
14	明 MING	日月南 ${\displaystyle \triangle RYS}$	明弦（RY日月）	明股（RS日南）	明勾（YS月南）
15	叀 ZHUAN	山川东 ${\displaystyle \triangle MSE}$	叀弦（MS山川）	叀股（ME山东）	叀勾（SE川东）

Close

In problems from Vol 2 to Vol 12, the names of these triangles are used in very terse terms. For instance

"明差","MING difference" refers to the "difference between the vertical side and horizontal side of MING triangle.

"叀差","ZHUANG difference" refers to the "difference between the vertical side and horizontal side of ZHUANG triangle."

"明差叀差并" means "the sum of MING difference and ZHUAN difference"${\displaystyle (b_{14}-a_{14})+(b_{15}-a_{15})}$

Length of Line Segments

This section (今问正数) lists the length of line segments, the sum and difference and their combinations in the diagram of round town, given that the radius r of inscribe circle is ${\displaystyle r=120}$ paces ${\displaystyle a_{1}=320}$,${\displaystyle b_{1}=640}$.

The 13 segments of ith triangle (i=1 to 15) are:

1. Hypoteneuse ${\displaystyle c_{i}}$
2. Horizontal ${\displaystyle a_{i}}$
3. Vertical ${\displaystyle b_{i}}$
4. :勾股和 :sum of horizontal and vertical ${\displaystyle a_{i}+b_{i}}$
5. :勾股校: difference of vertical and horizontal ${\displaystyle b_{i}-a_{i}}$
6. :勾弦和: sum of horizontal and hypotenuse ${\displaystyle a_{i}+c_{i}}$
7. :勾弦校: difference of hypotenuse and horizontal ${\displaystyle c_{i}-a_{i}}$
8. :股弦和: sum of hypotenuse and vertical ${\displaystyle b_{i}+c_{i}}$
9. :股弦校: difference of hypotenuse and vertical ${\displaystyle c_{i}-b_{i}}$
10. :弦校和: sum of the difference and the hypotenuse ${\displaystyle c_{i}+(b_{i}-a_{i})}$
11. :弦校校: difference of the hypotenuse and the difference ${\displaystyle c_{i}-(b_{i}-a_{i})}$
12. :弦和和: sum the hypotenuse and the sum of vertical and horizontal ${\displaystyle a_{i}+b_{i}+c_{i}}$
13. :弦和校: difference of the sum of horizontal and vertical with the hypotenuse ${\displaystyle a_{i}+b_{i}}$

Among the fifteen right angle triangles, there are two sets of identical triangles:

${\displaystyle \triangle TRD}$=${\displaystyle \triangle RMZ}$,

${\displaystyle \triangle YSG}$=${\displaystyle \triangle BLJ}$

that is

${\displaystyle a_{6}=a_{7}}$;

${\displaystyle b_{6}=b_{7}}$;

${\displaystyle c_{6}=c_{7}}$;

${\displaystyle a_{8}=a_{9}}$;

${\displaystyle b_{8}=b_{9}}$;

${\displaystyle c_{8}=c_{9}}$;

Segment numbers

There are 15 x 13 =195 terms, their values are shown in Table 1:^[2]

Definitions and formula

Tian yuan shu

From problem 14 onwards, Li Zhi introduced "Tian yuan one" as unknown variable, and set up two expressions according to Section Definition and formula, then equate these two tian yuan shu expressions. He then solved the problem and obtained the answer.

Question 14:"Suppose a man walking out from West gate and heading south for 480 paces and encountered a tree. He then walked out from the North gate heading east for 200 paces and saw the same tree. What is the radius of the round own?"

Algorithm: Set up the radius as Tian yuan one, place the counting rods representing southward 480 paces on the floor, subtract the tian yuan radius to obtain

：${\displaystyle 480-x}$

元

。

Then subtract tian yuan from eastward paces 200 to obtain:

${\displaystyle 200-x}$

元

multiply these two expressions to get：${\displaystyle x^{2}-680x+96000}$

元

元

that is${\displaystyle x^{2}-680x+96000=2x^{2}}$

thus：${\displaystyle -x^{2}-680x+96000=0}$

元

Solve the equation and obtain ${\displaystyle r=120}$

Problem 14

Given the sum of GAO difference and MING difference is 161 paces and the sum of MING difference and ZHUAN difference is 77 paces. What is the diameter of the round city?

Answer: 120 paces.

Algorithm:^[13]

Given

${\displaystyle (b_{7}-a_{7})+(b_{8}-a_{8})=161}$

${\displaystyle (b_{14}-a_{14})+(b_{15}-a_{15})=77}$

：Add these two items, and divide by 2; according to #Definitions and formula, this equals to HUANGJI difference:

${\displaystyle (b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}) \over 2}$ ${\displaystyle =(b_{12}-a_{12})}$

${\displaystyle b_{12}-a_{12}=}$${\displaystyle 161+77 \over 2}$${\displaystyle =119}$

Let Tian yuan one as the horizontal of SHANGPING (SG):

${\displaystyle x=a_{8}}$

${\displaystyle x+161}$ =${\displaystyle x+(b_{7}-a_{7})+(b_{8}-a_{8})=a_{8}+(b_{7}-a_{7})+(b_{8}-a_{8})}$

${\displaystyle =b_{7}}$ (#Definition and formula)

Since ${\displaystyle a_{8}+b_{7}=c_{12}}$ (Definition and formula)

${\displaystyle c_{12}=x+b_{7}=2*x+(b_{7}-a_{7})+(b_{8}-a_{8})=2*x+161}$

${\displaystyle c_{12}^{2}=(x+b_{7})^{2}=(2*x+161)^{2}=4*x^{2}+644*x+25921}$

${\displaystyle c_{12}^{2}-(b_{12}-a_{12})^{2}}$

${\displaystyle =4*x^{2}+644*x+25921-}$${\displaystyle ((b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}))^{2} \over 4}$

${\displaystyle =4*x^{2}+644*x+11760=d}$(diameter of round town),

${\displaystyle d^{2}=(4*x^{2}+644*x+11760)^{2}=16*x^{4}+5152*x^{3}+508816*x^{2}+15146880*x+138297600}$

Now, multiply the length of RZ by ${\displaystyle 4*x}$

${\displaystyle 4*x*b_{7}=4*x*(x+(b_{7}-a_{7})+(b_{8}-a_{8}))=4*x*(x+161)=4*x^{2}+644*x}$

multiply it with the square of RS:

${\displaystyle d^{2}=4*x*b_{7}*c_{12}^{2}=}$${\displaystyle (4*x^{2}+644*x)*(4*x^{2}+644*x+25921)=}$${\displaystyle 16*x^{4}+5152*x^{3}+518420*x^{2}+16693124}$

equate the expressions for the two ${\displaystyle d^{2}}$

thus

${\displaystyle 16*x^{4}+5152*x^{3}+518420*x^{2}+16693124=}$${\displaystyle 16*x^{4}+5152*x^{3}+508816*x^{2}+15146880*x+138297600}$

We obtain:

${\displaystyle 9604*x^{2}+1546244*x-138297600=0}$

solve it and we obtain ${\displaystyle x=a_{8}=64}$;

This matches the horizontal of SHANGPING 8th triangle in #Segment numbers.^[14]